The shortest distance between the airport A and airport B travels by airplane which departs airport A at a heading of 300 (N 60 W) is 401 miles.
When the two sides of and one angle is known, then to find the third side, the law of cosine is used.
It can be given as,
[tex]c^2=a^2+b^2-2ab\cos C\\a^2=c^2+b^2-2ab\cos A\\b^2=a^2+c^2-2ab\cos B[/tex]
Here, a,b and c are the sides of the triangle and A,B and C are the angles of the triangle.
An airplane departs airport A at a heading of 300 (N 60 W). After traveling 320 miles, the airplane adjusts its course to 350 (N 10 W) and flies an additional 112 miles to reach airport B.
The image of this problem is attached below. In this image, the two sides 112 miles and 320 miles are shown and the angle between them is 130 degrees. Thus, the value of x is,
[tex]x=\sqrt{112^2+130^2-2(112)(130)\cos 130^o}\\x=401\rm\; miles[/tex]
Hence, the shortest distance between the airport A and airport B travels by airplane which departs airport A at a heading of 300 (N 60 W) is 401 miles.
Learn more about the law of cosine here;
https://brainly.com/question/4372174
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Answer:
401 miles
Step-by-step explanation:
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